Can someone please verify whether my proof is correct?

Let $a,b\in G$ with $|a|=m$ and $|b|=n$ and $gcd(m,n)=1$. Show that $\left \langle a \right \rangle \cap \left \langle b \right \rangle = \left \{ e \right \}$.


By the fundamental theorem of cyclic groups, if $|a|=m$ and $|b|=n$, then the order of $\left \langle a \right \rangle \cap \left \langle b \right \rangle$ is a divisor of $m$ and $n$. Then $|\left \langle a \right \rangle \cap \left \langle b \right \rangle|=gcd(m,n)=1$. Additionally, $\left \langle a \right \rangle \cap \left \langle b \right \rangle$ has exactly one subgroup of order $1$, which is the trivial subgroup $\left \{ e \right \}$. However, since there can only be exactly one such subgroup of that order, it must be that $\left \langle a \right \rangle \cap \left \langle b \right \rangle = \left \{ e \right \}$. $\square$

  • $\begingroup$ I would say you could even end it after saying the order of the intersection must be one, because there is only one group of order one. $\endgroup$ – Dave Feb 5 '18 at 4:55

Yes, your answer is logically right. But I think it could be more clear and simple if there are some changes:

$\langle a\rangle\cap\langle b\rangle$ is a common subgroup of $\langle a\rangle$ and $\langle b\rangle$. According to the Lagrange Theorem, $|\langle a\rangle\cap\langle b\rangle |$ is a common divisor of both $|a|$ and $|b|$, namely $m$ and $n$. However, we have known that gcd$(m,n)=1$. Therefore, $|\langle a\rangle\cap\langle b\rangle |=1$.

It’s already sufficient.Problem solved.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.